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A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of number of successes.

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  • To check if a given distribution is a probability distribution of random variable, the sum of the individual probabilties should add up to 1 (i.e, $\sum P(X_i) = 1$). Also 0 $\lt$ P(X) $\leq$ 1.
Step 1:
Let $X$ denote the number of doublets.
Possible doublets are $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$
Clearly $X$ can take values as $0,1,2,3$ and 4
Probability of getting a doublet =$\large\frac{6}{36}=\frac{1}{6}$
Probability of not getting a doublet =$1-\large\frac{1}{36}=\frac{1}{6}$
Step 2:
Now $P(X=0)=P$(no doublets)
$\qquad\qquad\quad\;\;=\large\frac{5}{6}\times \frac{5}{6}\times\frac{5}{6}\times \frac{5}{6}$
Step 3:
$P(X=1)=P$(one doublet and three non-doublets)
$\qquad\quad\;\;=\large\frac{1}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}+\times \frac{5}{6}\times \frac{1}{6}\times \frac{5}{6}\times \frac{5}{6}+\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\times \frac{5}{6}+\times \frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}$
$\qquad\quad\;\;\;=4\big(\large\frac{1}{6}\times \frac{5^3}{6^3}\big)$
Step 4:
$P(X=2)=P$(two doublet and two non-doublets)
$\qquad\quad\;\;=\large\frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}\times \frac{5}{6}+\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}+\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}+\times \frac{1}{6}\times \frac{5}{6}\times \frac{1}{6}\times \frac{5}{6}$
$\qquad\quad\;\;\;=4\big(\large\frac{1}{6^2}\times \frac{5^2}{6^2}\big)$
Step 5:
$P(X=3)=P$(three doublets and one non-doublets)
$\qquad\quad\;\;=\large\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}+\times \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}\times \frac{1}{6}+\times \frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}+\times \frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}$
$\qquad\quad\;\;\;=4\big(\large\frac{1}{6^3}\times \frac{5}{6}\big)$
Step 6:
$P(X=4)=\large\frac{1}{6}\times \large\frac{1}{6}\times\large\frac{1}{6}\times\large\frac{1}{6}$
Thus the required probability distribution is
answered Sep 27, 2013 by sreemathi.v

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